I am learning about elliptic curves.

Consider the elliptic curve $y^2=x^3+2$ over $\mathbb Q$. Then $P=(-1,1)$ is a point on it.

I want to find its multiples i.e., $2P,3P, \cdots$

The addition of two points $P=(x_1,y_1),~Q=(x_2,y_2)$ is obtained by intersecting the line $\bar{PQ}$ intersecting the elliptic curve, i.e., if $R=(x_3,y_3)$ be the addition of $P$ and $Q$, then it is given by \begin{align} x_3&=\lambda^2-x_1-x_2 \\ y_3&=\lambda(x_1-x_3)-y_1, \\ \lambda&=\frac{y_2-y_1}{x_2-x_1} \end{align} But when we double a point, the slope $\lambda$ will be as follows: $$\lambda=\frac{3x_1^2+2}{2y_1}.$$ So in our case $\lambda=\frac{5}{2}$. Hence:

$$x_3=25/4+1+1=33/4, ~y_3=5/2(-1-33/4)-1=-193/8.$$

So $2P=(\frac{33}{4},-\frac{193}{8})$.

Is it correct ?

Is it the only way ?

Edit: I already got the answer as mentioned in the folloing comment sections. But I couldn't understand why my approach didn't work ? Are the formulas that I used correct ?

  • $\begingroup$ An interesting fact is that some elliptic curves have some special properties that make point doubling and addition "easier" (for example a montgomery curve). You can check here for some of them (en.wikipedia.org/wiki/…) $\endgroup$
    – Alex Them
    Commented Oct 3, 2022 at 9:55
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    $\begingroup$ Not sure, but it seems that $(33/4,-841/16)$ is not on the elliptic curve. You used $25/4$ instead of $5/2$… $\endgroup$
    – Aphelli
    Commented Oct 3, 2022 at 9:57
  • 1
    $\begingroup$ My answer here might be of help: math.stackexchange.com/questions/4158446/… $\endgroup$ Commented Oct 3, 2022 at 9:57
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    $\begingroup$ If you are learning about elliptic curves, you may want to use PARI/GP for computations. For example, in your case, try E=ellinit(ellfromeqn(-y^2+x^3+2)); P=[-1,1]; for(n=1,4,print(n," ",ellmul(E,P,n))). This will find $2P=(17/4,-71/8)$ and $3P,4P$. $\endgroup$
    – Somos
    Commented Oct 3, 2022 at 10:48
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    $\begingroup$ Here's the LMFDB entry for this specific elliptic curve: lmfdb.org/EllipticCurve/Q/1728/n/4 $\endgroup$ Commented Oct 3, 2022 at 17:11

1 Answer 1


Basically, your slope calculation was not correct.

Let $P=(-1,1)$

$$\lambda=\frac{3 x_1^2+2}{2y_1} = \frac{3\cdot 1^2+0}{2}= \color{red}{\frac{3}{2}}$$ The formula is

$$x_e = \lambda^2 - 2x_1$$

$$x_3=(9/4) - 2(-1) = 17/4$$

For $y_3$ formula is

$$y_3 = \lambda(x_1 − x_3 ) − y_1$$

\begin{align} y_3&=\left[(3/2)(-1-17/4)\right]-1\\ &=\left[(3/2)(-21/4)\right]-1\\ &=(-63/8)-1\\ &=(-63/8)-1\\ &=-71/8\\ \end{align}

Verifacation with SageMath on the SageMathCell

A = 0
B = 2
E = EllipticCurve([A,B])

P = E(-1,1)


(17/4 : -71/8 : 1)
  • 1
    $\begingroup$ Thanks a lot for pointing my error $\endgroup$
    – MAS
    Commented Oct 7, 2022 at 15:25

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